Find all matrices which satisfy $M^2-3M+3I = 0$ I am trying to find all matrices which solve the matrix equation 
$$M^2 -3M +3I=0$$
Since this doesn't factor I tried expanding this in terms of the coordinates of the matrix.  It also occurs to me to put it into "vertex" form:
$$M^2 - 3M + \frac{9}{4}I+\frac{3}{4}I=0$$
$$(M-\frac{3}{2}I)^2 = -\frac{3}{4}I$$
but this doesn't look much better.
What I found from expanding by coordinates was, if $M=\pmatrix{a & b \\ c & d}$ then 
$$\pmatrix{a^2+bc -3a + 3& ab + bd - 3b \\ ac+cd-3c & bc+d^2-3d+3} = \pmatrix{0&0\\0&0}$$
From the off-diagonal entries I get that either 
$$a+d-3=0$$
or 
$$b=c=0$$
If $a+d-3\not=0$ then $a^2-3a+3=0$ and likewise for $d$.  Then we get more cases for $a$ and $d$.
If $a+d-3=0$ the upper-left is unchanged and the lower-right is 
$$bc + (3-a)^2-3(3-a)+3 = 0$$ 
which simplifies to the same thing from the upper-left and so is redundant.  In the off-diagonals
$$ac+c(a-3)-3c = 0 \Rightarrow $$
$$2ac-6c = 0$$
We again get cases, and I suppose after chasing cases enough you get the solution set.
However, it just feels like this can't be the intended solution given how tedious and uninformative all of this case-chasing is.  Is there some bigger idea I'm missing?
 A: $m^2 - 3m + 3 = 0\\
\lambda = \frac {3}{2} \pm i\frac {\sqrt {3}}{2}$
You could say that it is all matrices with eigenvalues equal to $\frac {3}{2} + i\frac {\sqrt {3}}{2},\frac {3}{2} - i\frac {\sqrt {3}}{2}$
If we restrict our universe to real $2\times 2$ matrices.
Then it would be all matrices with characteristic equations equal to:
$\lambda^2 - 3\lambda + 3 = 0$
We are looking for matrices with trace equal to 3, and determinant 3.
$\begin{bmatrix} a & b\\ -\frac {a^2 -3a + 3}{b} & 3-a \end {bmatrix}$
A: Minimal polynomial of $M, m_M(x),$ is a factor of $x^2-3x+3=[x-(\frac{3+i\sqrt3}2)][x-(\frac{3-i\sqrt3}2)]$
Either $m_M(x)=x-(\frac{3+i\sqrt3}2)\implies M=[\frac{3+i\sqrt3}2]$
or $m_M(x)=x-(\frac{3-i\sqrt3}2)\implies M=[\frac{3-i\sqrt3}2]$
or $m_M(x)=x^2-3x+3\implies$ the eigenvalues of $M$ are $\frac{3\pm i\sqrt3}2$
In case of $2\times2$ matrices, product of eigenvalues $=\det(M)=3$, sum of eigenvalues $=\text{Tr}(M)=3$
We have $M=\begin{bmatrix}a&b\\c&3-a\end{bmatrix}$ and $3a-a^2-bc=3; a,b,c\in\Bbb C$. 
You could go for $3\times3,4\times4,...$ matrices by defining the same eigenvalues and conditions. In case you are looking for real matrices, you just have to take the real subset of these matrices.
A: You already found the "completion of the square"
$$
\left( {M - {3 \over 2}I} \right)^{\,2}  =  - {3 \over 4}I
$$
Then you can write
$$
\left( {i{2 \over {\sqrt 3 }}\left( {M - {3 \over 2}I} \right)} \right)^{\,2}  = X^{\,2}  = I
$$
So we are essentially looking for the square roots of the unit matrix, also complex,
or for the square roots of $- \, I$.
You can find various papers dealing with this subject, for example this related post
or this thesis.
-- p.s. --
I thought you were interested in the general case of $n \times n$ matrices.
If your question is limited to $2 \times 2$ then the $\sqrt{\pm I}$ is easily found on the net (e.g.,see the hint on Pauli matrices).
A: hint
By Cayley-Hamilton,
if the caracteristic polynom is 
$$x^2-3x+3$$
then
$$M^2-3M+3I=0$$
then
$$(a-x)(d-x)-bc=x^2-3x+3$$
